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ELECTROCHEMISTRY II

• ELECTROLYTIC CELLS• PREDICTION OF PRODUCTS• FARADAY’S LAW• Chapter 21.4-21.7 Silberberg

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Chapter 21

Electrochemistry: Chemical Change and Electrical Work

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Electrochemistry: Chemical Change and Electrical Work

21.4 Free Energy and Electrical Work

21.5 Electrochemical Processes in Batteries

21.6 Corrosion: An Environmental Voltaic Cell

21.7 Electrolytic Cells: Using Electrical Energy to Drive Nonspontaneous Reactions

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Goals & Objectives

• See the following Learning Objectives on pages 969-970.

• Understand these Concepts:• 21.13-16; 19-23.

• Master these Skills:• 21.5-13.

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Products of Electrolysis

Electrolysis is the splitting (lysing) of a substance by the input of electrical energy.

During electrolysis of a pure, molten salt, the cation will be reduced and the anion will be oxidized.

During electrolysis of a mixture of molten salts- the more easily oxidized species (stronger reducing agent) reacts

at the anode, and- the more easily reduced species (stronger oxidizing agent) reacts

at the cathode.

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Sample Problem 21.8 Predicting the Electrolysis Products of a Molten Salt Mixture

PROBLEM: A chemical engineer melts a naturally occurring mixture of NaBr and MgCl2 and decomposes it in an electrolytic cell. Predict the substance formed at each electrode, and write balanced half-reactions and the overall cell reaction.

PLAN: We need to determine which metal and nonmetal will form more easily at the electrodes. We list the ions as oxidizing or reducing agents.

If a metal holds its electrons more tightly than another, it has a higher ionization energy (IE). Its cation will gain electrons more easily, and it will be the stronger oxidizing agent.

If a nonmetal holds its electrons less tightly than another, it has a lower electronegativity (EN). Its anion will lose electrons more easily, and it will be the reducing agent.

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Sample Problem 21.8

SOLUTION:

Possible oxidizing agents: Na+, Mg2+

Possible reducing agents: Br-, Cl-

Mg is to the right of Na in Period 3. IE increases from left to right across the period, so Mg has the higher IE and gives up its electrons less easily. The Mg2+ ion has a greater attraction for e- than the Na+ ion.

Br is below Cl in Group 7A. EN decreases down the group, so Br accepts e- less readily than Cl. The Br- ion will lose its e- more easily, so it is more easily oxidized.

Mg2+(l) + 2e- → Mg(l) [cathode; reduction]

2Br-(l) → Br2(g) + 2e- [anode; oxidation]

The overall cell reaction is: Mg2+(l) + 2Br-(l) → Mg(l) + Br2(g)

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Electrolytic Cells

• The Electrolysis of Molten Potassium Chloride– produces liquid potassium at one electrode– produces gaseous chlorine at the other

• Diagram this cell using inert electrodes– write the electrode reactions– label the electrodes– indicate the direction of electron flow

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Electrolysis of Aqueous Salt Solutions

When an aqueous salt solution is electrolyzed- The strongest oxidizing agent (most positive electrode potential) is

reduced, and- The strongest reducing agent (most negative electrode potential) is

oxidized.

Overvoltage is the additional voltage needed (above that predicted by E° values) to produce gases at metal electrodes.

Overvoltage needs to be taken into account when predicting the products of electrolysis for aqueous solutions.Overvoltage is 0.4 – 0.6 V for H2(g) or O2(g).

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• Cations of less active metals (Au, Ag, Cu, Cr, Pt, Cd) are reduced to the metal.

• Cations of more active metals are not reduced. H2O is reduced instead.

• Anions that are oxidized, because of overvoltage from O2 formation, include the halides, except for F-.

• Anions that are not oxidized include F- and common oxoanions. H2O is oxidized instead.

Summary of the Electrolysis of Aqueous Salt Solutions

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Sample Problem 21.9 Predicting the Electrolysis Products of Aqueous Salt Solutions

PROBLEM: What products form at which electrode during electrolysis of aqueous solution of the following salts? (a) KBr (b) AgNO3 (c) MgSO4

PLAN: We identify the reacting ions and compare their electrode potentials with those of water, taking the 0.4 – 0.6 V overvoltage into account. The reduction half-reaction with the less negative E° occurs at the cathode, while the oxidation half-reaction with the less positive E° occurs at the anode.

Despite the overvoltage, which makes E for the reduction of water between -0.8 and -1.0 V, H2O is still easier to reduce than K+, so H2(g) forms at the cathode.

SOLUTION:

(a) KBr K+(aq) + e- → K(s) E° = -2.93 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V

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(b) AgNO3 Ag+(aq) + e- → Ag(s) E° = 0.80 V 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V

Sample Problem 21.9

2Br-(aq) → Br2(l) + 2e- E° = 1.07 V 2H2O(l) → O2(g) + 4H+(aq) + 4e- E° = 0.82 V

The overvoltage makes E for the oxidation of water between 1.2 and 1.4 V. Br- is therefore easier to oxidize than water, so Br2(g) forms at the anode.

As the cation of an inactive metal, Ag+ is a better oxidizing agent than H2O, so Ag(s) forms at the cathode.

NO3- cannot be oxidized, because N is already in its highest (+5)

oxidation state. Thus O2(g) forms at the anode:

2H2O(l) → O2(g) + 4H+(aq) + 4e-

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(c) MgSO4 Mg2+(aq) + 2e- → Mg(s) E° = -2.37 V 2H2O(l) + 2e- → H2(g) + 2OH-(aq) E° = -0.42V

Sample Problem 21.9

Mg2+ is a much weaker oxidizing agent than H2O, so H2(g) forms at the cathode.

SO42- cannot be oxidized, because S is already in its highest (+6)

oxidation state. Thus O2(g) forms at the anode:

2H2O(l) → O2(g) + 4H+(aq) + 4e-

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Electrolytic Cells

• The Electrolysis of Aqueous Potassium Chloride Solution– produces gaseous hydrogen at one electrode

and the solution becomes basic at this electrode– produces gaseous chlorine at the other


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Electrolytic Cells• The Electrolysis of Aqueous Potassium

Sulfate Solution– produces hydrogen at one electrode and the

solution becomes basic at this electrode– produces gaseous oxygen at the other and the

solution becomes acidic at this electrode


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Electrode Products

• Generalization:– In all electrolytic cells the most easily

reduced species is reduced and the most easily oxidized species is oxidized.

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Electrode Products

• Prediction of Electrode Products• Anode

– Cl-, Br-, I- -> Cl2, Br2, I2 respectively

– all other anions produce O2

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Electrode Products

• Prediction of Electrode Products• Cathode

– uses the Activity Series or Electromotive Series– neutral solutions:

• above Zn -> H2

• Zn and below -> metal

– acid solutions:• H and above -> H2

• below H -> metal

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Activity Series

• A Partial Activity Series of the Elements– Li– Na See page 132 in McMurry– Al– Zn– Cr See page 171 in Silberberg – Ni 5th edition– H– Ag See page 165 in Silberberg– Pt 6th edition– Au

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Electrode Products

• Predict the products expected for each of the following electrolytic cells:

• Anode Cathode• NaBr(molten)

– Na lies above Zn

• MgI2(aq)– Mg lies above Zn

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Electrode Products

• Predict the products expected for each of the following electrolytic cells:

• Anode Cathode• ZnSO4(aq)

• Ni(NO3)(aq,H+)– Ni lies above H

• CuSO4(aq)– Cu lies below Zn

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Figure 21.28 The electrolysis of water.

Oxidation half-reaction2H2O(l) → 4H+(aq) + O2(g) + 4e-

Reduction half-reaction2H2O(l) + 4e- → 2H2(g) + 2OH-(aq)

Overall (cell) reaction2H2O(l) → H2(g) + O2(g)

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Table 21.3 Some Ions Measured with Ion-Specific Electrodes

Species Detected Typical Sample

NH3/NH4+ Industrial wastewater, seawater

CO2/HCO3- Blood, groundwater

F- Drinking water, urine, soil, industrial stack gases

Br- Grain, plant tissue

I- Milk, pharmaceuticals

NO3- Soil, fertilizer, drinking water

K+ Blood serum, soil, wine

H+ Laboratory solutions, soil, natural waters

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Figure 21.14 Minimocroanalysis.

A microelectrode records electrical impulses of a single neuron in a monkey’s visual cortex. The electrical potential of a nerve cell is due to the difference in concentration of [Na+] and [K+] ions inside and outside the cell.

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Faraday’s Law of Electrolysis

• The amount of substance undergoing chemical reaction at each electrode during electrolysis is directly proportional to the amount of electricity passing through the electrolytic cell.

• The unit faraday refers to the amount of electricity that reduces one equivalent of a species at the cathode and oxidizes one equivalent at the anode.

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Michael Faraday

• Michael Faraday, (September 21, – August 25, 1867) was an English chemist and physicist who contributed to the fields of electromagnetism and electrochemistry..

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Andre- Marie Ampere

• André-Marie Ampère (January 20, 1775 – June 10, 1836), was a French physicist who is generally credited as one of the main discoverers of electromagnetism. The SI unit of measurement of electric current, the ampere, is named after him

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Stoichiometry of Electrolysis

Faraday’s law of electrolysis states that the amount of substance produced at each electrode is directly proportional to the quantity of charge flowing through the cell.

The current flowing through the cell is the amount of charge per unit time. Current is measured in amperes.

Current x time = charge

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Figure 21.29 A summary diagram for the stoichiometry of electrolysis.

MASS (g)of substance oxidized or

reduced

AMOUNT (mol)of substance oxidized or

reduced

CHARGE(C)

CURRENT (A)

AMOUNT (mol)of electrons transferred

M (g/mol)

balanced half-reaction

Faraday constant (C/mol e-)

time (s)

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Sample Problem 21.10 Applying the Relationship Among Current, Time, and Amount of Substance

PROBLEM: A technician plates a faucet with 0.86 g of Cr metal by electrolysis of aqueous Cr2(SO4)3. If 12.5 min is allowed for the plating, what current is needed?

PLAN: To find the current, we divide charge by time, so we need to find the charge. We write the half-reaction for Cr3+ reduction to get the amount (mol) of e- transferred per mole of Cr. We convert mass of Cr needed to amount (mol) of Cr. We can then use the Faraday constant to find charge and current.

divide by M

mass (g) of Cr needed

mol of Cr

3 mol e- = 1 mol Cr

divide by time in s

mol e- transferred Charge (C)

1 mol e- = 9.65x104 C

current (A)

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Sample Problem 21.10

SOLUTION:Cr3+(aq) + 3e- → Cr(s)

0.86 g Cr x 1 mol Cr

52.00 g Crx 3 mol e-

1 mol Cr= 0.050 mol e-

Charge (C) = 0.050 mol e- x 9.65x104 C

1 mol e- = 4.8x103 C

Current (A) =charge (C)

time (s)= 4.8x103 C

12.5 minx 1 min

60 s= 6.4 C/s = 6.4 A

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• A faraday corresponds to the gain or loss of one mole of electrons.

• Thus– one faraday = 6.022x1023 electrons– one faraday = one mole of electrons

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• A coulomb is defined as the amount of charge that passes a given point when a current of one ampere(amp) flows for one second.– one coulomb = amp x sec

• Or– one amp = one Coulomb per second(C/s)

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• A faraday is equivalent to 96,487 coulombs or more frequently 9.65x104C– one faraday = one mole of electron– one faraday = 9.65x104C– 9.65x104C = one mole of electrons

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• Determine the mass of palladium produced by the reduction of palladium(II) ions to palladium metal during the passage of 3.20 amperes of current through a solution of palladium(II) sulfate for 30.0 minutes.

• Determine the volume of oxygen (measured at STP) produced in this cell.

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• Determine the products, and amounts of each expected, when an acidic solution of NiSO4 is electrolyzed by a current of 5.46amps for 2.00 hours. Ni lies above H on the activity series.

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• Determination of the charge on an ion by electrolysis– An aqueous solution of a palladium salt was

electrolyzed for 1.00 hours with a current of 1.50amperes. This produced 2.977g of Pd metal at the cathode. Determine the charge on the palladium ion in this salt.

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Corrosion is the process whereby metals are oxidized to their oxides and sulfides.

Corrosion: an Environmental Voltaic Cell

The rusting of iron is a common form of corrosion.

- Rusting requires moisture, and occurs more quickly at low pH, in ionic solutions, and when the iron is in contact with a less active metal.

- Rust is not a direct product of the reaction between Fe and O2, but arises through a complex electrochemical process.

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The Rusting of Iron

Fe(s) → Fe2+(aq) + 2e- [anodic region; oxidation]O2(g) + 4H+(aq) + 4e- → 2H2O(l) [cathodic region; reduction]

The loss of iron:

2Fe(s) + O2(g) + 4H+(aq) → 2Fe2+(aq) + 2H2O(l) [overall]

The rusting process:

2Fe2+(aq) + ½O2(g) + (2 + n)H2O(l) → Fe2O3·nH2O(s) + 4H+(aq)

Overall reaction:

H+ ions are consumed in the first step, so lowering the pH increases the overall rate of the process. H+ ions act as a catalyst, since they are regenerated in the second part of the process.

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Figure 21.22 The corrosion of iron.

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Figure 21.23 Enhanced corrosion at sea.

The high ion concentration of seawater enhances the corrosion of iron in hulls and anchors.

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Figure 21.24 The effect of metal-metal contact on the corrosion of iron.

Fe in contact with Cu corrodes faster.

Fe in contact with Zn does not corrode. The process is known as cathodic protection.

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Figure 21.25 The use of sacrificial anodes to prevent iron corrosion.

In cathodic protection, an active metal, such as zinc, magnesium, or aluminum, acts as the anode and is sacrificed instead of the iron.

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Electrolytic Cells

An electrolytic cell uses electrical energy from an external source to drive a nonspontaneous redox reaction.

Cu(s) → Cu2+(aq) + 2e- [anode; oxidation]Sn2+(aq) + 2e- → Sn(s) [cathode; reduction]

Cu(s) + Sn2+(aq) → Cu2+(aq) + Sn(s) E°cell = -0.48 V and ΔG° = 93 kJ

As with a voltaic cell, oxidation occurs at the anode and reduction takes place at the cathode.

An external source supplies the cathode with electrons, which is negative, and removes then from the anode, which is positive. Electrons flow from cathode to anode.

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Figure 21.26 The tin-copper reaction as the basis of a voltaic and an electrolytic cell.

voltaic cell

Sn(s) → Sn2+(aq) + 2e-

Cu2+(aq) + 2e- → Cu(s)

Cu2+(aq) + Sn(s) → Cu(s) + Sn2+(aq)

electrolytic cell

Cu(s) → Cu2+(aq) + 2e-

Sn2+(aq) + 2e- → Sn(s)

Sn2+(aq) + Cu(s) → Sn(s) + Cu2+(aq)

Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

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Figure 21.27 The processes occurring during the discharge and recharge of a lead-acid battery.

VOLTAIC (discharge)

ELECTROLYTIC (recharge)

Switch

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Table 21.4 Comparison of Voltaic and Electrolytic Cells

Cell Type DG Ecell

Electrode

Name Process Sign

Voltaic

Voltaic

< 0

< 0

> 0

> 0

Anode

Cathode

Oxidation

Reduction

-

+

Electrolytic

Electrolytic

> 0

> 0

< 0

< 0

Anode

Cathode

Oxidation

Reduction -

+

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Zinc Air Hearing Aid Batteries

• Zinc–air batteries are electro-chemical batteries powered by oxidizing zinc with oxygen from the air. These batteries have high energy densities and are relatively inexpensive to produce. Sizes include very small button cells for hearing aids

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Zinc Air Hearing Aid Batteries

• Here are the chemical equations for the

zinc–air cell:• Anode: Zn + 4OH– → Zn(OH)4

2– + 2e–

• Cathode: 1/2 O2 + H2O + 2e– → 2OH–

• Overall: 2Zn + O2 → 2ZnO (E0 = 1.59 V)

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